One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end fo another horizontal thin copper wire of lenth L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is

Options

A. 0.25
B.  0.50
C.  2.00
D. `4.00`

Explanation

To solve the problem, we need to find the ratio of the elongation in the thin wire to that in the thick wire when both are subjected to the same force. We will use the concept of Young's modulus, which relates stress and strain in materials.

### Step-by-Step Solution:

1. **Understanding Young's Modulus:**
Young's modulus (Y) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
where:
- \( F \) = force applied,
- \( A \) = cross-sectional area,
- \( \Delta L \) = change in length (elongation),
- \( L \) = original length.

2. **Cross-Sectional Areas:**
- For the thick wire (length = 2L, radius = 2R):
\[ A_{\text{thick}} = \pi (2R)^2 = 4\pi R^2 \]
- For the thin wire (length = L, radius = R):
\[ A_{\text{thin}} = \pi R^2 \]

3. **Applying Young's Modulus for Each Wire:**
- For the thick wire:
\[ Y = \frac{F}{A_{\text{thick}}} \cdot \frac{L}{\Delta L_1} \implies Y = \frac{F}{4\pi R^2} \cdot \frac{2L}{\Delta L_1} \]
- For the thin wire:
\[ Y = \frac{F}{A_{\text{thin}}} \cdot \frac{L}{\Delta L_2} \implies Y = \frac{F}{\pi R^2} \cdot \frac{L}{\Delta L_2} \]

4. **Equating Young's Modulus:**
Since both wires are made of the same material (copper), their Young's moduli are equal:
\[ \frac{F \cdot 2L}{4\pi R^2 \cdot \Delta L_1} = \frac{F \cdot L}{\pi R^2 \cdot \Delta L_2} \]

5. **Canceling Common Terms:**
We can cancel \( F \) and \( \pi R^2 \) from both sides:
\[ \frac{2L}{4 \Delta L_1} = \frac{L}{\Delta L_2} \]

6. **Rearranging the Equation:**
Rearranging gives:
\[ \Delta L_2 = \frac{L \cdot 4 \Delta L_1}{2L} = 2 \Delta L_1 \]

7. **Finding the Ratio of Elongations:**
The ratio of elongation in the thin wire to that in the thick wire is:
\[ \frac{\Delta L_2}{\Delta L_1} = 2 \]

### Final Answer:
The ratio of the elongation in the thin wire to that in the thick wire is \( 2 \).

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